Let f:[-1,2] rightarrow R be defined by f(x)= | vedclass

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(C) The function is $f(x) = [x^2 - 3]$.We know that the greatest integer function $[g(x)]$ is discontinuous at points where $g(x)$ is an integer.Here,

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$3$

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(C) The function is $f(x) = [x^2 - 3]$.We know that the greatest integer function $[g(x)]$ is discontinuous at points where $g(x)$ is an integer.Here, $g(x) = x^2 - 3$.For $x \in (-1, 2)$, the range of $g(x) = x^2 - 3$ is:When $x = -1$, $g(x) = (-1)^2 - 3 = -2$.When $x = 0$, $g(x) = 0^2 - 3 = -3$.When $x = 2$, $g(x) = 2^2 - 3 = 1$.So, for $x \in (-1, 2)$, $g(x)$ takes values in the interval $(-3, 1)$.The integer values that $g(x)$ takes in this interval are $\{-2, -1, 0\}$.We need to find the values of $x$ such that $x^2 - 3 = k$ for $k \in \{-2, -1, 0\}$.$1$) $x^2 - 3 = -2 \implies x^2 = 1 \implies x = \pm 1$. Since we are looking in $(-1, 2)$, only $x = 1$ is in the interval.$2$) $x^2 - 3 = -1 \implies x^2 = 2 \implies x = \pm \sqrt{2}$. Since we are looking in $(-1, 2)$, only $x = \sqrt{2}$ is in the interval.$3$) $x^2 - 3 = 0 \implies x^2 = 3 \implies x = \pm \sqrt{3}$. Since we are looking in $(-1, 2)$, only $x = \sqrt{3}$ is in the interval.Thus, the points of discontinuity are $x \in \{1, \sqrt{2}, \sqrt{3}\}$.The total number of points of discontinuity is $3$.

Let $f:[-1,2] \rightarrow R$ be defined by $f(x)=[x^2-3]$ where $[\cdot]$ denotes the greatest integer function. The number of points of discontinuity for the function $f$ in the interval $(-1,2)$ is:

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